Question

Heya!

Answer 1

Answer:

the answer is in last picture

Answer 2

Answer:

See Below.

Step-by-step explanation:

We are given that PQRS is a square. A, B, C, and D are the points of the sides PQ, QR, RS, and SP, respectively.

And we are given that AQ = BR = CS = DP.

This is shown in the figure below.

And we want to prove that ABCD is a square.

Since PQRS is a square, it follows that:

$m\angle P=m\angle Q=m\angle R=m\angle S=90^\circ$

Likewise:

$PQ=QR=RS=SP$

PQ is the sum of the segments PA and AQ:

$PQ=PA+AQ$

Likewise, QR is the sum of the segments QB and BR:

$QR=QB+BR$

Since AQ = BR and PQ = QR:

$PA+AQ=QB+AQ$

Therefore:

$PA=QB$

Likewise, RS is the sum of the segments RC and CS:

$RS=RC+CS$

Since RS = PQ and CS = AQ:

$RC+CS=PA+CS$

Thus:

$RC=PA=QB$

Repeating this procedure for the remaining side, we acquire that:

$PA=QB=RC=SD$

And since each of the angles ∠P, ∠Q, ∠R, and ∠S is 90°, by the SAS Theorem, we acquire:

$\Delta AQB\cong\Delta BRC\cong \Delta CSD\cong \Delta DPA$

Then by CPCTC, we acquire:

$AB=BC=CD=DA$

However, this means that ABCD could be either a rhombus or a square, so we need to prove that the angles are right angles.

The interior angles of a triangle always sum to 180°. Since they are right triangles, the two other angles must equal 90°. Thus, for ΔAQB:

$m\angle QAB+m\angle QBA=90^\circ$

By CPCTC, ∠PAD≅∠QBA. Thus:

$m\angle QAB+m\angle PAD=90^\circ$

∠DAB forms a linear pair. Therefore:

$m\angle PAD+m\angle QAB+m\angle DAB=180$

By substitution and simplification:

$m\angle DAB=90^\circ$

So, ∠DAB is a right angle.

By repeating this procedure, we can establish that ∠ABC, ∠BCD, and ∠CDA are all right angles. This is not necessary, however. Since we concluded that AB = BC = CD = DA, and that we have one right angle, then ABCD must be a square.